Question

the sum of the first three terms of a sequence Is 42 The fourth term is 24. find u1 and the common difference

Answer 1

$10=2d$

Given:

The sum of the first three terms of a sequence Is 42.

The fourth term is 24.

To find:

The  u₁ and the common difference

Step-by-step explanation:

nth term of a sequence is

$u_n=u_1+(n-1)d$

where, u₁ is first term and d is common difference.

4th term is 24.

$u_4=u_1+(4-1)d$

$24=u_1+3d$           ...(i)

Sum of nth term of an AP is

$S_n=\dfrac{n}{2}[2u_1+(n-1)d]$

Sum of first three terms is 42.

$S_3=\dfrac{3}{2}[2u_1+(3-1)d]$

$42=\dfrac{3}{2}[2u_1+2d]$

Divide both sides by 3.

$14=u_1+d$          ...(ii)

On subtract (ii) from (i), we get

$10=2d$

$5=d$

Put d=5 in equation (ii).

$14=u_1+5$

$14-5=u_1$

$9=u_1$

Therefore, the first term is $u_1=9$ and common difference is $d=5$ .